= Solution
For a right prism with equilateral triangular ends, the triangle area is $\sqrt3\alpha^2/4$. Thus its volume and external area are
$$
V=\frac{\sqrt3}{4}\alpha^2\beta,\qquad
A=\frac{\sqrt3}{2}\alpha^2+3\alpha\beta.
$$
Apply the <Lagrange multiplier> method to $V-\lambda(A-A_0)$ with positive lengths. The stationarity equations are
$$
\frac{\sqrt3}{2}\alpha\beta=\lambda(\sqrt3\alpha+3\beta),
\qquad
\frac{\sqrt3}{4}\alpha^2=3\lambda\alpha.
$$
The second gives $\lambda=\sqrt3\alpha/12$. Substitution in the first then gives $6\beta=\sqrt3\alpha+3\beta$, or $\beta=\alpha/\sqrt3$. The area constraint becomes
$$
\frac{3\sqrt3}{2}\alpha^2=A_0=\frac92\,\mathrm{m^2}.
$$
Consequently
$$
\boxed{\alpha=3^{1/4}\,\mathrm m,\qquad
\beta=3^{-1/4}\,\mathrm m.}
$$
The resulting volume is $V_{\max}=3^{3/4}/4\,\mathrm{m^3}$. As a check on the maximum, eliminating $\beta$ gives $V(\alpha)=\sqrt3A_0\alpha/12-\alpha^3/8$, whose second <derivative> is negative for every admissible positive $\alpha$; volume also tends to zero at both degenerate boundary shapes.
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